I want to prove that for any sets $A,B,C$: $$\left(A \subseteq B \land \vert A \rvert = \lvert A \cup C \rvert\right) \implies \vert B \rvert = \lvert B \cup C \rvert$$ We have $A \subseteq B \implies \vert A \rvert \le \lvert B \rvert$, so $\vert A \cup C \rvert \le \lvert B \rvert$. Also, $B \subseteq (B \cup C) \implies \vert B \rvert \le \lvert B \cup C \rvert$, so it would suffice to show $\vert B \rvert \ge \lvert B \cup C \rvert$. This is where I get stuck. How do I proceed from here?
Show $\left(A \subseteq B \land \vert A \rvert = \lvert A \cup C \rvert\right) \implies \vert B \rvert = \lvert B \cup C \rvert$
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elementary-set-theory
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2I would try and figure out when $|A| = |A \cup C|$ rather than work with inequalities, personally. – 2017-02-01
1 Answers
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Since $|A| = |A \cup C|$, we know there is a one-to-one mapping between $A$ and $A \cup C$; let's say this is mapping $f$
We also know that $A \subseteq B$, so consider all the elements in $B-A$: these are exactly the same elements that are in $(B\cup C) - (A \cup C)$. So, define your one-to-one mapping $f'$ between $B$ and $B \cup C$ as follows:
For any $x \in A$: $f'(a) = f(a)$
For any $x \in B - A$: $f'(a) = a$
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0Does the lemma follow from the following facts? $\lvert X \cup Y \rvert = \lvert X \rvert +\lvert Y \rvert-\lvert X \cap Y \rvert$, so $\lvert X \cup Y \rvert = \lvert X \rvert$ gives $\lvert X \cap Y \rvert = \lvert Y \rvert$ and there cannot be a bijection between $Y$ and $X \cap Y$ unless $Y \subseteq X$? – 2017-02-01
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1@Zelazny That would work, sure! – 2017-02-01
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0@Zelazny Oh hell! I just realized I was thinking only of finite sets! For infinite sets, the Lemma doesn't even hold! OK, let me redo my Answer. Hold on! – 2017-02-01
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0@Zelazny OK, I redid my answer! – 2017-02-01